Concerning the definitions of the enthalpy, the Helmholtz free energy and the Gibbs free energy of a systemWhy is the Gibbs Free Energy $F-HM$?Gibbs free energy and maximum workHelmholtz Free Energy vs Gibbs Free Energy in Landau TheoryGibbs' free energy and Helmholtz free energyHow could chemical potential be interpreted as the molar Gibbs free energy?Gibbs Free Energy of Two concurring PhasesHelmholtz Free Energy at EquilibriumPhysical Significance of $U$ (Internal Energy ) , $H$ (Enthalpy) , $F$ (Free Energy) and $G$ (Gibbs Free Energy)?Gibbs free energy the 3 equation confusionIs Entropy a monotonically increasing function of Gibbs Free Energy/ Helmholtz free energy/ Enthalpy?
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Concerning the definitions of the enthalpy, the Helmholtz free energy and the Gibbs free energy of a system
Why is the Gibbs Free Energy $F-HM$?Gibbs free energy and maximum workHelmholtz Free Energy vs Gibbs Free Energy in Landau TheoryGibbs' free energy and Helmholtz free energyHow could chemical potential be interpreted as the molar Gibbs free energy?Gibbs Free Energy of Two concurring PhasesHelmholtz Free Energy at EquilibriumPhysical Significance of $U$ (Internal Energy ) , $H$ (Enthalpy) , $F$ (Free Energy) and $G$ (Gibbs Free Energy)?Gibbs free energy the 3 equation confusionIs Entropy a monotonically increasing function of Gibbs Free Energy/ Helmholtz free energy/ Enthalpy?
$begingroup$
The definition of enthalpy, $$H = U + PV,$$ assumes that the system is in a constant-pressure environment. Similarly, the definition of the Helmholtz free energy of the system, $$F = U - TS,$$ assumes a constant-temperature environment. The definition of the Gibbs free energy, $$G = U + PV - TS,$$ takes for granted both of the aformentioned assumptions. Does this mean, for example, that the enthalpy of a system is undefined for a system with a volume-dependent pressure? I have similar questions about the Helmholtz free energy and the Gibbs free energy.
thermodynamics
asked 7 hours ago
PiKindOfGuyPiKindOfGuy
589622
$endgroup$
add a comment |
$begingroup$
The definition of enthalpy, $$H = U + PV,$$ assumes that the system is in a constant-pressure environment. Similarly, the definition of the Helmholtz free energy of the system, $$F = U - TS,$$ assumes a constant-temperature environment. The definition of the Gibbs free energy, $$G = U + PV - TS,$$ takes for granted both of the aformentioned assumptions. Does this mean, for example, that the enthalpy of a system is undefined for a system with a volume-dependent pressure? I have similar questions about the Helmholtz free energy and the Gibbs free energy.
thermodynamics
asked 7 hours ago
PiKindOfGuyPiKindOfGuy
589622
$endgroup$
$begingroup$
I can understand where your puzzle comes from. Some textbook (such as Schroeder's book, section 5.1), for the convenience of presenting, uses phrases like constant pressure or constant temperature. But it doesn't exclude other conditions.
$endgroup$
– user115350
48 mins ago
add a comment |
$begingroup$
The definition of enthalpy, $$H = U + PV,$$ assumes that the system is in a constant-pressure environment. Similarly, the definition of the Helmholtz free energy of the system, $$F = U - TS,$$ assumes a constant-temperature environment. The definition of the Gibbs free energy, $$G = U + PV - TS,$$ takes for granted both of the aformentioned assumptions. Does this mean, for example, that the enthalpy of a system is undefined for a system with a volume-dependent pressure? I have similar questions about the Helmholtz free energy and the Gibbs free energy.
thermodynamics
asked 7 hours ago
PiKindOfGuyPiKindOfGuy
589622
$endgroup$
The definition of enthalpy, $$H = U + PV,$$ assumes that the system is in a constant-pressure environment. Similarly, the definition of the Helmholtz free energy of the system, $$F = U - TS,$$ assumes a constant-temperature environment. The definition of the Gibbs free energy, $$G = U + PV - TS,$$ takes for granted both of the aformentioned assumptions. Does this mean, for example, that the enthalpy of a system is undefined for a system with a volume-dependent pressure? I have similar questions about the Helmholtz free energy and the Gibbs free energy.
thermodynamics
thermodynamics
asked 7 hours ago
PiKindOfGuyPiKindOfGuy
589622
asked 7 hours ago
PiKindOfGuyPiKindOfGuy
589622
asked 7 hours ago
PiKindOfGuyPiKindOfGuy
589622
asked 7 hours ago
PiKindOfGuyPiKindOfGuy
589622
asked 7 hours ago
PiKindOfGuyPiKindOfGuy
589622
589622
$begingroup$
I can understand where your puzzle comes from. Some textbook (such as Schroeder's book, section 5.1), for the convenience of presenting, uses phrases like constant pressure or constant temperature. But it doesn't exclude other conditions.
$endgroup$
– user115350
48 mins ago
add a comment |
$begingroup$
I can understand where your puzzle comes from. Some textbook (such as Schroeder's book, section 5.1), for the convenience of presenting, uses phrases like constant pressure or constant temperature. But it doesn't exclude other conditions.
$endgroup$
– user115350
48 mins ago
$begingroup$
I can understand where your puzzle comes from. Some textbook (such as Schroeder's book, section 5.1), for the convenience of presenting, uses phrases like constant pressure or constant temperature. But it doesn't exclude other conditions.
$endgroup$
– user115350
48 mins ago
$begingroup$
I can understand where your puzzle comes from. Some textbook (such as Schroeder's book, section 5.1), for the convenience of presenting, uses phrases like constant pressure or constant temperature. But it doesn't exclude other conditions.
$endgroup$
– user115350
48 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Those are the definitions of each. They don't assume anything about the system and can always be applied. You are getting mixed up with the scenarios in which they are usually applied since nice things happen. For example, for a system at constant pressure (and number of particles) $Delta H=Q$, where $Q$ is the heat that enters or leaves the system.
To add some more detail, this can be seen by substituting in the thermodynamic identity
$$text dU=Ttext dS-Ptext dV+mutext d N$$
into the differential of one of your thermodynamic potentials. For example, as mentioned above we have
$$text dH=text dU+Ptext dV+Vtext dP$$
so then
$$text dH=Ttext dS+Vtext dP+mutext dN$$
i.e. at constant pressure and number of particles $text dH=Ttext dS=text dQ$
You also say that the Gibbs free energy takes both mentioned assumptions "for granted", but see what happens if you do this process with the Gibbs free energy at constant temperature and pressure. It is a very important relation.
These processes are more generally called Legendre transformations
answered 7 hours ago
Aaron StevensAaron Stevens
13.9k42252
$endgroup$
add a comment |
$begingroup$
I agree with @Aaron Stevens. There are no built in assumptions of constant pressure or temperature in these definitions.
The Hemlholtz free energy, Gibbs free energy, Enthalpy and Internal Energy are sometimes referred to as thermodynamic potentials. For a discussion of these check out
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/helmholtz.html
Hope this helps.
answered 5 hours ago
Bob DBob D
4,3932318
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Those are the definitions of each. They don't assume anything about the system and can always be applied. You are getting mixed up with the scenarios in which they are usually applied since nice things happen. For example, for a system at constant pressure (and number of particles) $Delta H=Q$, where $Q$ is the heat that enters or leaves the system.
To add some more detail, this can be seen by substituting in the thermodynamic identity
$$text dU=Ttext dS-Ptext dV+mutext d N$$
into the differential of one of your thermodynamic potentials. For example, as mentioned above we have
$$text dH=text dU+Ptext dV+Vtext dP$$
so then
$$text dH=Ttext dS+Vtext dP+mutext dN$$
i.e. at constant pressure and number of particles $text dH=Ttext dS=text dQ$
You also say that the Gibbs free energy takes both mentioned assumptions "for granted", but see what happens if you do this process with the Gibbs free energy at constant temperature and pressure. It is a very important relation.
These processes are more generally called Legendre transformations
answered 7 hours ago
Aaron StevensAaron Stevens
13.9k42252
$endgroup$
add a comment |
$begingroup$
Those are the definitions of each. They don't assume anything about the system and can always be applied. You are getting mixed up with the scenarios in which they are usually applied since nice things happen. For example, for a system at constant pressure (and number of particles) $Delta H=Q$, where $Q$ is the heat that enters or leaves the system.
To add some more detail, this can be seen by substituting in the thermodynamic identity
$$text dU=Ttext dS-Ptext dV+mutext d N$$
into the differential of one of your thermodynamic potentials. For example, as mentioned above we have
$$text dH=text dU+Ptext dV+Vtext dP$$
so then
$$text dH=Ttext dS+Vtext dP+mutext dN$$
i.e. at constant pressure and number of particles $text dH=Ttext dS=text dQ$
You also say that the Gibbs free energy takes both mentioned assumptions "for granted", but see what happens if you do this process with the Gibbs free energy at constant temperature and pressure. It is a very important relation.
These processes are more generally called Legendre transformations
answered 7 hours ago
Aaron StevensAaron Stevens
13.9k42252
$endgroup$
add a comment |
$begingroup$
Those are the definitions of each. They don't assume anything about the system and can always be applied. You are getting mixed up with the scenarios in which they are usually applied since nice things happen. For example, for a system at constant pressure (and number of particles) $Delta H=Q$, where $Q$ is the heat that enters or leaves the system.
To add some more detail, this can be seen by substituting in the thermodynamic identity
$$text dU=Ttext dS-Ptext dV+mutext d N$$
into the differential of one of your thermodynamic potentials. For example, as mentioned above we have
$$text dH=text dU+Ptext dV+Vtext dP$$
so then
$$text dH=Ttext dS+Vtext dP+mutext dN$$
i.e. at constant pressure and number of particles $text dH=Ttext dS=text dQ$
You also say that the Gibbs free energy takes both mentioned assumptions "for granted", but see what happens if you do this process with the Gibbs free energy at constant temperature and pressure. It is a very important relation.
These processes are more generally called Legendre transformations
answered 7 hours ago
Aaron StevensAaron Stevens
13.9k42252
$endgroup$
Those are the definitions of each. They don't assume anything about the system and can always be applied. You are getting mixed up with the scenarios in which they are usually applied since nice things happen. For example, for a system at constant pressure (and number of particles) $Delta H=Q$, where $Q$ is the heat that enters or leaves the system.
To add some more detail, this can be seen by substituting in the thermodynamic identity
$$text dU=Ttext dS-Ptext dV+mutext d N$$
into the differential of one of your thermodynamic potentials. For example, as mentioned above we have
$$text dH=text dU+Ptext dV+Vtext dP$$
so then
$$text dH=Ttext dS+Vtext dP+mutext dN$$
i.e. at constant pressure and number of particles $text dH=Ttext dS=text dQ$
You also say that the Gibbs free energy takes both mentioned assumptions "for granted", but see what happens if you do this process with the Gibbs free energy at constant temperature and pressure. It is a very important relation.
These processes are more generally called Legendre transformations
answered 7 hours ago
Aaron StevensAaron Stevens
13.9k42252
edited 4 hours ago
answered 7 hours ago
Aaron StevensAaron Stevens
13.9k42252
answered 7 hours ago
Aaron StevensAaron Stevens
13.9k42252
answered 7 hours ago
Aaron StevensAaron Stevens
13.9k42252
13.9k42252
add a comment |
add a comment |
$begingroup$
I agree with @Aaron Stevens. There are no built in assumptions of constant pressure or temperature in these definitions.
The Hemlholtz free energy, Gibbs free energy, Enthalpy and Internal Energy are sometimes referred to as thermodynamic potentials. For a discussion of these check out
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/helmholtz.html
Hope this helps.
answered 5 hours ago
Bob DBob D
4,3932318
$endgroup$
add a comment |
$begingroup$
I agree with @Aaron Stevens. There are no built in assumptions of constant pressure or temperature in these definitions.
The Hemlholtz free energy, Gibbs free energy, Enthalpy and Internal Energy are sometimes referred to as thermodynamic potentials. For a discussion of these check out
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/helmholtz.html
Hope this helps.
answered 5 hours ago
Bob DBob D
4,3932318
$endgroup$
add a comment |
$begingroup$
I agree with @Aaron Stevens. There are no built in assumptions of constant pressure or temperature in these definitions.
The Hemlholtz free energy, Gibbs free energy, Enthalpy and Internal Energy are sometimes referred to as thermodynamic potentials. For a discussion of these check out
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/helmholtz.html
Hope this helps.
answered 5 hours ago
Bob DBob D
4,3932318
$endgroup$
I agree with @Aaron Stevens. There are no built in assumptions of constant pressure or temperature in these definitions.
The Hemlholtz free energy, Gibbs free energy, Enthalpy and Internal Energy are sometimes referred to as thermodynamic potentials. For a discussion of these check out
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/helmholtz.html
Hope this helps.
answered 5 hours ago
Bob DBob D
4,3932318
answered 5 hours ago
Bob DBob D
4,3932318
answered 5 hours ago
Bob DBob D
4,3932318
answered 5 hours ago
Bob DBob D
4,3932318
4,3932318
add a comment |
add a comment |
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$begingroup$
I can understand where your puzzle comes from. Some textbook (such as Schroeder's book, section 5.1), for the convenience of presenting, uses phrases like constant pressure or constant temperature. But it doesn't exclude other conditions.
$endgroup$
– user115350
48 mins ago